The height of a rectangle is increasing at a rate of $11$ centimeters per hour and the width of the rectangle is decreasing at a rate of $9$ centimeters per hour. At a certain instant, the height is $3$ centimeters and the width is $8$ centimeters. What is the rate of change of the area of the rectangle at that instant (in square centimeters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $-99$ (Choice B) B $-61$ (Choice C) C $99$ (Choice D) D $61$
Solution: Setting up the math Let... $h(t)$ denote the height of the rectangle at time $t$, $w(t)$ denote the width of the rectangle at time $t$, and $A(t)$ denote the area of the rectangle at time $t$. We are given that $h'(t)=11$ and $w'(t)=-9$ (notice that $w'$ is negative). We are also given that that $h(t_0)=3$ and $w(t_0)=8$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures The measures relate to each other through the formula for the area of a rectangle: $A(t)=h(t)w(t)$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=h'(t)w(t)+h(t)w'(t)$ Using the information to solve Let's plug ${h'(t_0)}={11}$, ${w(t_0)}={8}$, ${h(t_0)}={3}$, and $C{w'(t_0)}=C{-9}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&={h'(t_0)}{w(t_0)}+{h(t_0)}C{w'(t_0)} \\\\ &=({11})({8})+({3})(C{-9}) \\\\ &=61 \end{aligned}$ In conclusion, the rate of change of the area of the rectangle at that instant is $61$ square centimeters per hour. Since the rate of change is positive, we know that the area is increasing.